Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.3 Derivatives Of Inverse Functions; Derivatives And Integrals Involving Exponential Functions - Exercises Set 6.3 - Page 432: 57

Answer

$$3$$

Work Step by Step

$$\eqalign{ & \mathop {\lim }\limits_{x \to 0} \frac{{{e^{3x}} - 1}}{x} \cr & {\text{the limit can be written as }} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{{e^{3x}} - {e^{3\left( 0 \right)}}}}{{x - 0}} \cr & {\text{from the definition of the derivative }}f'\left( c \right) = \mathop {\lim }\limits_{x \to c} \frac{{f\left( x \right) - f\left( c \right)}}{{x - c}},{\text{ we compare}} \cr & {\text{that }}\mathop {\lim }\limits_{x \to 0} \frac{{{e^{3x}} - {e^{3\left( 0 \right)}}}}{{x - 0}} \Rightarrow \,\,f\left( x \right) = {e^{3x}}{\text{ and }}c = 0,{\text{ Thus}}{\text{,}} \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {{e^{3x}}} \right] \cr & f'\left( x \right) = 3{e^{3x}} \cr & and \cr & f'\left( 0 \right) = 3{e^{3\left( 0 \right)}} \cr & f'\left( 0 \right) = 3 \cr & {\text{Then}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{{e^{3x}} - 1}}{x}{\text{ }} = 3 \cr} $$
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